Question

In the given figure LM || AB, AL = x − 3, AC = 2x, BM = x − 2 and BC = 2x + 3 find the value of x.

Answer 1

hello mate here is answer for you !
there is two triangles CAB and CLM and they are similar since the sides AB and LM are parallel .so from the property of similarity .
CL is similar to CA and CM is similar to CB .
so CL/CA=CM/CB...(eq.1)

CL=CA-AL=2x-(x-3)=x+3
and CM=CB-BM=2x+3-(x-2) =x+5.

now you put the value above in eq.1
we get ,(x+3)/2x=(x+5)/(x-2)

now solve it for value x(Cross multiply)
the equation came out is .
x^2+9x+6=0
now solve this quadratic equation for x.
use formula ,x=-b+-√b^2-4ac/2a.
put a=1,b=9 and c=6

x=-9+-√9^2-4×1×6/2×1

so ..

x=-9+-√57/2.
here both values would be -ve so ,take the greater one !
take the greater value as value of x.
or ,it's x=-9+√57/2 .ans.
it took about 2 hours for me to thing about this puzzle !
but at last I did it anyhow, really it was a great learning for me as well.

so thank you for posting this puzzle!
hope it helped you !

Answer 2

Given :

In ∆ABC , LM//AB

AL = x - 3 , AC = 2x ,

BM = x - 2 , and BC = 2x + 3 ,

Solution :

In ∆ABC ,LM// AB

=> AL/LC = BM/MC

[ Thales theorem ]

( x - 3 )/[ 2x - ( x - 3 )] = (x-2)/[ (2x+3)-(x-2)]

=> (x-3)/(x+3) = ( x-2 )/(x+5)

=> (x-3)(x+5) = ( x-2)(x+3)

=> x² + 2x - 15 = x² + x - 6

=> 2x - 15 = x - 6

=> 2x - x = 15 - 6

x = 9

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